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Triginometry , algebra help

  • Thread starter tweety1234
  • Start date
1. Homework Statement

In △ABC, AB=(2-x) cm, BC=(x+1) cm and ∠ABC=120°:

(a) Show that AC^2=x^2-x+7.

(b) Find the value of x for which AC has a minimum value


3. The Attempt at a Solution

I am having trouble with part 'b'



am suppose to complete the square , but I dont seem to get the right answer.

[tex] x^2 -x +7 =0 [/tex]

[tex] x^2 -x = -7 [/tex]

[tex] (x- \frac{1}{2} )^2 = -7 + \frac{1}{4} [/tex]

[tex](x- \frac{1}{2} )^2 = \frac{-27}{4} [/tex]

[tex] x- \frac{1}{2} = -\frac{\sqrt27}{2}[/tex]


and than am I suppose to solve for x ? what does it mean by minimum value?
 

tiny-tim

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uhh? :confused: where did "=0" come from??

just minimise (x - 1/2)2 + 27/4 :wink:
I dont understand what you mean by 'minimise' ? [tex] \frac{1}{4} + \frac{27}{4} = 7 [/tex]

x=7?
 

tiny-tim

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I dont understand what you mean by 'minimise' ?
Find the value of x for which (x - 1/2)2 + 27/4 is a minimum … what is that minimum? :smile:
 
Find the value of x for which (x - 1/2)2 + 27/4 is a minimum … what is that minimum? :smile:

IS the minimum 27/4 , when x = 1/2?
 

tiny-tim

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Mentallic

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[tex](x- \frac{1}{2} )^2 = \frac{-27}{4} [/tex]

[tex] x- \frac{1}{2} = -\frac{\sqrt27}{2}[/tex]


This step is not correct as taking the square root of a negative number (right hand side of equation) results in complex, imaginary numbers; not the negative square root as you've done. Since x is imaginary, it means that [tex]x^2-x+7 \neq 0[/tex]
And of course, you've already seen that the minimum value is 27/4 which is more than 0.
 

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